Question Number 210261 by klipto last updated on 04/Aug/24
$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{\mathrm{sinAcosA}}−\boldsymbol{\mathrm{sinBcosB}}}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{B}}}=\boldsymbol{\mathrm{tan}}\left(\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}\right) \\ $$
Answered by efronzo1 last updated on 04/Aug/24
$$\:\:\:\frac{\mathrm{sin}\:\mathrm{A}\:\mathrm{cos}\:\mathrm{A}−\mathrm{sin}\:\mathrm{B}\:\mathrm{cos}\:\mathrm{B}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{A}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{B}}\:\overset{?} {=}\:\mathrm{tan}\:\left(\mathrm{A}−\mathrm{B}\right) \\ $$$$\:\:\:\:\underbrace{\Subset} \\ $$
Answered by A5T last updated on 04/Aug/24
$${tan}\left({A}−{B}\right)=\frac{{sin}\left({A}−{B}\right)}{{cos}\left({A}−{B}\right)}=\frac{{sinAcosA}−{sinBcosB}}{{cos}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {B}} \\ $$